The coefficient of earth pressure at rest is given by (μ = Poisson's ratio):
Try it first: pick an option, then read the explanation.
Asked in SSC JE 2010
Detailed Explanation
At rest, the soil behind a rigid, unmoving wall cannot strain sideways. Using Hooke's law in three dimensions with zero horizontal strain gives the ratio of horizontal to vertical stress.
\[ \varepsilon_x = \frac{1}{E}\left[\sigma_x - \mu(\sigma_y + \sigma_z)\right] = 0, \quad \sigma_x = \sigma_y \]
\[ \sigma_x(1-\mu) = \mu\sigma_z \quad\Rightarrow\quad K_0 = \frac{\sigma_x}{\sigma_z} = \frac{\mu}{1-\mu} \]
Why “μ/(1 − μ)” Is Correct
Setting the lateral strain to zero and the two horizontal stresses equal gives K0 = μ/(1 − μ). For μ = 0.33, K0 ≈ 0.5.
| Coefficient | Formula (cohesionless soil) | φ = 30° |
|---|---|---|
| Active, Ka | (1 − sin φ)/(1 + sin φ) = tan²(45° − φ/2) | 1/3 ≈ 0.333 |
| At rest, K0 | μ/(1 − μ) (elastic) or ≈ 1 − sin φ (Jaky) | ≈ 0.5 |
| Passive, Kp | (1 + sin φ)/(1 − sin φ) = tan²(45° + φ/2) | 3 |
For Rankine's theory with a level backfill, Ka × Kp = 1 and Ka < K0 < Kp.
Why the Other Options Are Wrong
- A. μ²/(1 − μ²): μ²/(1 − μ²) does not come from the zero-strain condition.
- C. (1 − μ)/μ: (1 − μ)/μ is the inverse of K₀ and would be greater than 1 for normal soils.
- D. (1 − μ²)/μ²: (1 − μ²)/μ² is not a valid expression for K₀.
Key Point
K₀ = μ/(1 − μ) (elastic); empirically K₀ ≈ 1 − sin φ′ (Jaky) for NC soils.
Key Concepts for Students
- Basement walls and bridge abutments that cannot move are designed for K₀.
- Overconsolidated soils have higher K₀, sometimes greater than 1.
- Ka < K₀ < Kp.
Practice more: this is question 22 in Earth Pressure and Retaining Wall MCQs – Part 2. Try the whole set and check every answer instantly.